LABEX 4.1 Xử lý tín hiệu số

Copy from figure windows and paste... The plot of the gain response of the stable transfer function HBPz obtained using< Insert MATLAB figures here.. Copy from figure windows and paste..

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LINEAR, TIMEINVARIANT DISCRETETIME SYSTEMS:

FREQUENCYDOMAIN REPRESENTATIONS

4.1 TRANSFER FUNCTION AND FREQUENCY RESPONSE

Project 4.1 Transfer Function Analysis

Answers:

Q4.1

The modified Program P3_1 to compute and plot the magnitude and phase spectra of a

% Program P3_1

% Evaluation of the DTFT

clf;

% Compute the frequency samples

of the DTFT

w = 0:8*pi/511:2*pi;

m = 5;

num =ones(1,m)/m;

h = freqz(num, 1, w);

% Plot the DTFT

subplot(2,1,1)

plot(w/pi,abs(h));grid title('Magnitude Spectrum | H(e^{j\omega})|')

xlabel('\omega /\pi');

ylabel('Amplitude');

subplot(2,1,2) plot(w/pi,angle(h));grid title('Phase Spectrum arg[H(e^{j\omega})]') xlabel('\omega /\pi');

ylabel('Phase in radians');

0

0.5

j )|

 / 

4

2

0

2

j )]

 / 

The types of symmetries exhibited by the magnitude and phase spectra are due to

1

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The results of Question Q2.1 can now be explained as follo -

0

0.5

j )|

2

1

0

1

j )]

0

0.5

j )|

 / 

4

2

0

2

j )]

 / 

nhau

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I shall choose the filter of Question Q4.2 for the following reason Vì đáp ng pha ứ

t t h n ố ơ

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

2

4

6

8 Magnitude Spectrum |H(e

j )|

 /

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1

0.5

0

0.5

1 Phase Spectrum arg[H(e

j )]

 /

Questions 4.2 and 4.3 obtained using the program developed in Question Q3.50

0 10 20 30 40 50 60 70 80 90 100

1.2

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8 impule

 /

3

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0 10 20 30 40 50 60 70 80 90 100

4

3

2

1

0

1

2

3

4

5

 / 

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

Real Part

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

Real Part

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4.2 TYPES OF TRANSFER FUNCTIONS

Project 4.2 Filters

% Program P4_1

% Impulse Response of Truncated Ideal Lowpass Filter

clf;

fc = 0.25;

n = [-6.5:1:6.5];

y = 2*fc*sinc(2*fc*n);k = n+6.5;

stem(k,y);title('N = 13');axis([0 13 -0.2 0.6]);

xlabel('Time index n');ylabel('Amplitude');grid;

Answers:

0.2

0.1

0

0.1

0.2

0.3

0.4

0.5

Time index n

[-6.5:1:6.5];

response of the FIR lowpass filter of Project 4.2 with a length of 20 and a cutoff

% Program P4_1

% Impulse Response of Truncated Ideal Lowpass Filter

clf;

fc = 0.45;

n = [-10:1:10];

y = 2*fc*sinc(2*fc*n);k = n+10;

stem(k,y);title('N = 20');axis([0 20 -0.2

1]);

xlabel('Time index

n');ylabel('Amplitude');grid;

5

0.2 0 0.2 0.4 0.6 0.8

N = 20

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The plot generated by running the modified program is given below:

response of the FIR lowpass filter of Project 4.2 with a length of 15 and a cutoff

% Program P4_1

% Impulse Response of Truncated Ideal

Lowpass Filter

clf;

fc = 0.65;

n = [-7.5:1:7.5];

y = 2*fc*sinc(2*fc*n);k = n+7.5;

stem(k,y);title('N = 15');axis([0 20

-0.2 0.6]);

xlabel('Time index

The plot generated by running the

Q4.10 The MATLAB program to compute and plot the amplitude response of the FIR

% Program P4_1

% Impulse Response of Truncated Ideal

Lowpass Filter

clf;

fc = 0.25;

n = [-6.5:1:6.5];

y = 2*fc*sinc(2*fc*n);k = n+6.5;

plot(k,y);title('N = 13');axis([0 13 -0.2

0.6]);

xlabel('Time index

Plots of the amplitude response of the

N = 15

0.2

0.1 0 0.1 0.2 0.3 0.4 0.5

Time index n

0.2

0.1 0 0.1 0.2 0.3 0.4 0.5

Time index n

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0 2 4 6 8 10 12

0.2

0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time index n

N= 19

0.2

0.1

0

0.1

0.2

0.3

0.4

0.5

0.6 N = 19

Time index n

% Program P4_2

% Gain Response of a Moving

Average Lowpass Filter

clf;

M = 2;

num = ones(1,M)/M;

[g,w] = gain(num,1);

plot(w/pi,g);grid axis([0 1 -50 0.5]) xlabel('\omega /\pi');ylabel('Gain in dB');

title(['M = ', num2str(M)])

Answers:

Q4.11 A plot of the gain response of a length-2 moving average filter obtained using

7

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

40

35

30

25

20

15

10

5

0

 / 

Q4.12 The required modifications to Program P4_2 to compute and plot the gain response

% Program P4_2

% Gain Response of a Moving Average Lowpass Filter

clf;

M = 2;

num =[1 1]/M;

plot(w/pi,g);grid

axis([0 1 -50 0.5])

xlabel('\omega /\pi');ylabel('Gain in dB');

The plot of the gain response for a cascade of 3 sections obtained using the

50

45

40

35

30

25

20

15

10

5

0

 / 

M = 3

Q4.13 The required modifications to Program P4_2 to compute and plot the gain response

% Program P4_2

% Gain Response of a Moving Average Lowpass Filter

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M = 5;

num =[1 -1 1 -1 1]/M;

plot(w/pi,g);grid

axis([0 1 -50 0.5])

xlabel('\omega /\pi');ylabel('Gain in dB');

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

40

35

30

25

20

15

10

5

0

M = 5

Q4.14 From Eq (4.16) for a 3-dB cutoff frequency c at 0.45 we obtain  = 0.0787

function of the first-order IIR lowpass and highpass filters, respectively, given by

HLP(z) =

HHP(z) =

9

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

40

35

30

25

20

15

10

5 0

 /

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

40

35

30

25

20

15

10

5

0

 /

A plot of the magnitude response of

From this plot we observe that the two filters

A plot of the sum of the

< Insert MATLAB figure(s) here Copy from figure window(s) and paste >

Q4.15 From Eq (4.24), we get substituting K = 10, B = 1.86

of 10 IIR lowpass filters as

1 z1

1 –  z1



10

 (7.85+7.85z1)/(2+13.7z1)

0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21

0.6

0.5

0.4

0.3

0.2

0.1 0 0.1

 / 

M = 5

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Using this value of  in Eq (4.15) we arrive at the transfer function of a first-order IIR lowpass filter

HLP,1(z)  1 – 

1  z1

1 –  z1  (7.7+7.7z1)/(2+13.4z1)

< Insert

Q4.16 Substituting o = 0.61 in Eq (4.19) we get  cos(0.61) 

transfer function of the IIR bandpass transfer function

HBP,1(z) =

transfer function of the IIR bandpass transfer function

HBP,2(z) =

11

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

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35

30

25

20

15

10

5 0

 /

M = 21

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

50

45

40

35

30

25

20

15

10

5 0

 /

HHP

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The plot of the gain response of the stable transfer function HBP(z) obtained using

of a stable IIR bandstop filter as

MATLAB is shown below:

From these plots we observe that the designed filters do/do not meet the

Q4.17 The transfer function of a comb filter derived from the prototype FIR lowpass filter

of Eq (4.38) is given by

G(z) = H0(zL) = Plots of the magnitude response of the above comb filter for the following values of

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From these plots we observe that the comb filter has _ notches at k = _

Q4.18 The transfer function of a comb filter derived from the prototype FIR highpass filter

G(z) = H1(zL) = Plots of the magnitude response of the above comb filter for the following values of

_

Q4.19 A copy of Program P4_3 is given below:

% Program P4_3

% Zero Locations of Linear Phase FIR Filters

clf;

b = [1 -8.5 30.5 -63];

num1 = [b 81 fliplr(b)];

num2 = [b 81 81 fliplr(b)];

num3 = [b 0 -fliplr(b)];

num4 = [b 81 -81 -fliplr(b)];

n1 = 0:length(num1)-1;

subplot(2,2,1); stem(n1,num1);

xlabel('Time index n');ylabel('Amplitude'); grid;

title('Type 1 FIR Filter');

pause

subplot(2,2,1); zplane(num1,1);

13

0 2 4 6 8

100

50 0 50 100

Time index n

Type 1 FIR Filter

100

50 0 50 100

Time index n

Type 2 FIR Filter

0 2 4 6 8

100

50 0 50 100

Time index n

Type 3 FIR Filter

100

50 0 50 100

Time index n

Type 4 FIR Filter

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disp('Zeros of Type 1 FIR Filter are');

disp(roots(num1));

disp(roots(num2));

disp(roots(num3));

disp(roots(num4));

- Các tín hiệu kế tiếp nhau cho ta thấy các giá trị M của tín hiệu là khác nhau : tín

hiệu 1 và 3 thì giá trị M chẵn và tín hiệu 2 và 4 thì giá trị M lẽ.

- Tín hiệu 1 có M=8 và a = 1.979

- Tín hiệu 2 có M= 9 và a = 1

- Tín hiệu 4 có M = 9 và a = 1

The plots of the impulse responses of the four FIR filters generated by running

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Filter #1 has zeros at z =

Plots of the phase response of each of these filters obtained using MATLAB are

Q4.20 The plots of the impulse responses of the four FIR filters generated by running

% Program P4_3

% Zero Locations of Linear Phase FIR Filters

clf;

b = [1 -8.5 30.5 -63];

num1 = [b 81 fliplr(b)];

num2 = [b 81 81 fliplr(b)];

num3 = [b 0 -fliplr(b)];

num4 = [b 81 -81 -fliplr(b)];

xlabel('Time index

n');ylabel('Amplitude'); grid;

xlabel('Time index

n');ylabel('Amplitude'); grid;

pause

15

0 2 4 6 8

100

50 0 50 100

Time index n

Type 1 FIR Filter

100

50 0 50 100

Time index n

Type 2 FIR Filter

0 2 4 6 8

100

50 0 50 100

Time index n

Type 3 FIR Filter

100

50 0 50 100

Time index n

Type 4 FIR Filter

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disp(roots(num1));

disp(roots(num2));

disp(roots(num3));

disp(roots(num4));

- Các tín hiệu kế tiếp nhau cho ta thấy các giá trị M của tín hiệu là khác nhau : tín hiệu 1 và 3 thì giá trị M chẵn và tín hiệu 2 và 4 thì giá trị M lẽ.

- Tín hiệu 1 có M=8 và a = 1.979

- Tín hiệu 4 có M = 9 và a = 1

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